Question

A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile strength of 5 pounds per millimeter (lbs/mm). The company tests a random sample of 42 such valve plates from a lot of 650 valve plates. The sample mean is a tensile strength of 5.0611 lbs/mm, and the population standard deviation is 0.2803 lbs/mm. Use α = .10 and test to determine whether the lot of valve plates has an average tensile strength of 5 lbs/mm.

Answer 1

Answer:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$    

$p_v =2*P(z>1.413)=0.158$  

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation  

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample  

$n=42$ sample size  

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

$p_v$ represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 5, the system of hypothesis would be:  

Null hypothesis:$\mu = 5$  

Alternative hypothesis:$\mu \neq 5$  

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$    

P-value

Since is a two sided test the p value would be:  

$p_v =2*P(z>1.413)=0.158$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance